Re: Mod 2011



On Thu, 18 Jan 2007 20:17:01 +0000 (UTC), Erick Bryce Wong wrote:
Dave Seaman <dseaman@xxxxxxxxxxxx> wrote:
Erick Bryce Wong wrote:
Jyrki Lahtonen <lahtonen@xxxxxx> wrote:
"Jyrki Lahtonen" <lahtonen@xxxxxx> wrote in message
news:eoihr0$1ac0$2@xxxxxxxxxxxxxxxxx
León-Sotelo wrote:
Determine the remainder when 1004! is divided by 2011.

In that case Wilson's theorem says that 2010! is congruent
to minus 1 modulo 2011. The rest is easy.

How does one easily resolve the issue of whether 1005! is +1 or -1?

Why would you need to?

Well, no reason other than to justify a claim such as "The rest is easy"...

At the time I asked that, I thought I saw a shortcut. Alas, I neglected to
write it in the margin.


--
Dave Seaman
U.S. Court of Appeals to review three issues
concerning case of Mumia Abu-Jamal.
<http://www.mumia2000.org/>
.



Relevant Pages

  • Re: Mod 2011
    ... to minus 1 modulo 2011. ... Dave Seaman ...
    (sci.math)
  • Re: Mod 2011
    ... Dave Seaman wrote: ... to minus 1 modulo 2011. ...
    (sci.math)
  • Re: Mod 2011
    ... to minus 1 modulo 2011. ... How does one easily resolve the issue of whether 1005! ...
    (sci.math)
  • Re: Mod 2011
    ... and work in the ring of integers modulo q. ... The sequence always starts: ... What's the difference between primes in sequences and above? ... for those primes congruent 1 mod 3. ...
    (sci.math)
  • Re: Mod 2011
    ... and work in the ring of integers modulo q. ... The sequence always starts: ... What's the difference between primes in sequences and above? ... for those primes congruent 1 mod 3. ...
    (sci.math)

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